Problem 106 - A Complete Resource Book in Chemistry for JEE Main

Question

In an organic compound of molar mass \(108 \mathrm{~g} \mathrm{~mol}^{-1} \mathrm{C}\), \(\mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 3.5\) by weight. The molecular formula of the compound can be (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{12} \mathrm{~N}_{2}\)

Step-by-Step Solution

Answer

Option (b) \(\text{C}_6\text{H}_8\text{N}_2\) is the molecular formula.

Step 1: Calculate the Molar Mass Ratio

Let's assign the molar mass of carbon (C), hydrogen (H), and nitrogen (N) as 12 g/mol, 1 g/mol, and 14 g/mol respectively. Given the ratios of the elements present are 9:1:3.5 by weight. For 9 parts by weight of C, 1 part of H, and 3.5 parts of N, let's calculate their masses:- Carbon: \[\frac{9}{(9 + 1 + 3.5)} \times 108 = 72 \text{ g/mol}\]- Hydrogen: \[\frac{1}{(9 + 1 + 3.5)} \times 108 = 8 \text{ g/mol}\]- Nitrogen: \[\frac{3.5}{(9 + 1 + 3.5)} \times 108 = 28 \text{ g/mol}\]

Step 2: Calculate the Number of Atoms

Now, calculate the moles of each of these atoms in the compound from their weights.- For Carbon: \[\text{Moles of } C = \frac{72}{12} = 6\]- For Hydrogen: \[\text{Moles of } H = \frac{8}{1} = 8\]- For Nitrogen: \[\text{Moles of } N = \frac{28}{14} = 2\]

Step 3: Find Molecular Formula

The empirical formula is derived from the smallest whole number ratio of moles calculated: C: H: N = 6:8:2. This can be simplified to C: H: N = 3:4:1. Hence the empirical formula is \(\text{C}_3\text{H}_4\text{N}\). Now calculate its empirical formula mass:- Empirical mass = \(3 \times 12 + 4 \times 1 + 1 \times 14 = 54\) g/mol.- To find the molecular formula, divide the given molar mass (108) by the empirical formula mass (54):\[\frac{108}{54} = 2\]Thus, the molecular formula = 2 \(\times\) (C₃H₄N) = \(\text{C}_6\text{H}_8\text{N}_2\).

Step 4: Compare and Choose

Compare the derived molecular formula \(\text{C}_6\text{H}_8\text{N}_2\) with the given options. Option (b) \(\text{C}_6\text{H}_8\text{N}_2\) matches our derived formula.

Key Concepts

Empirical Formula Calculation

Calculating the empirical formula is a fundamental step in determining the molecular formula of an organic compound. The empirical formula represents the simplest whole-number ratio of elements in a compound. It involves a series of straightforward calculations to identify how many moles of each element are present based on their weight and molar mass.

To do this, we start with the weight or percentage composition of each element in the compound. The given compound has a weight ratio of carbon (C), hydrogen (H), and nitrogen (N) as 9:1:3.5. To find the moles, divide the weight contributions of each element by their respective molar mass:

• Carbon: Divide the weight by 12 g/mol, the molar mass of carbon.
• Hydrogen: Divide the weight by 1 g/mol.
• Nitrogen: Divide the weight by 14 g/mol.

After calculating the moles of each element, determine the simplest ratio among them to arrive at the empirical formula. In this exercise, the empirical formula is calculated as \( \text{C}_3\text{H}_4\text{N} \), derived from the simplified mole ratio of 3:4:1.

Moles and Ratios

Understanding moles and ratios is essential in translating elemental compositions into meaningful chemical formulas. Moles allow us to express amounts of a substance in terms of the number of entities (atoms, molecules), rather than mass.

After determining the individual weights of elements in their given ratio, convert these weights into moles by using molar masses:

• For carbon, divide the calculated molar weight by 12 g/mol.
• For hydrogen, use its molar mass of 1 g/mol.
• For nitrogen, divide by 14 g/mol.

This conversion gives us the number of moles of each element in the compound.

The next step is to express these moles as ratios. To achieve this, divide each mole value by the smallest number of moles calculated. This will simplify the data into a whole-number ratio, giving us insight into the proportion of each element within the compound. Ultimately, these ratios help identify the empirical formula accurately.

Organic Chemistry

Organic chemistry is the study of carbon-containing compounds, which encompasses a vast array of substances from simple molecules like methane to complex biomolecules like DNA. Central to organic chemistry is the concept of molecular formulas, which provide precise insight into the types and numbers of atoms present in a molecule.

In the context of determining molecular formulas, it is essential to comprehend the relationship between the empirical formula, molar mass, and how they interact to provide a complete picture of the compound.

• Begin with an empirical formula that offers the simplest atomic ratio.
• Use the molar mass as a benchmark to find how many empirical formula units are present within the actual molecule.
• Multiply the empirical formula by this factor to get the true molecular formula.

In this exercise, the given organic compound's molecular formula is finally determined from the empirical formula \( \text{C}_3\text{H}_4\text{N} \), adjusting by a factor of 2 derived from the molar mass to reach \( \text{C}_6\text{H}_8\text{N}_2 \). This process is a key application of organic chemistry, providing both the structure and identity of an unknown compound.